Introductio in analysin infinitorum

NUMERORUM. 273 $(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})(1+x^{32})$ &c., CAP. XVI. in qua Exponentes ipsius $x$ in ratione dupla progrediuntur. Hæc expressio si evolvatur, reperietur quidem hæc Series $1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 + x^8 +$ &c.; quoniam vero dubium esse potest, utrum hæc Series in infinitum hac lege geometrica progrediatur, hanc ipsam Seriem investigemus. Sit igitur $P = (1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})$ &c., ac ponatur Series per evolutionem oriunda $P = 1 + \alpha x + \beta x^2 + \gamma x^3 + \delta x^4 + \epsilon x^5 + \zeta x^6 + \eta x^7 + \theta x^8 +$ &c., Patet autem si loco $x$ scribatur $xx$, tum prodire productum $(1+xx)(1+x^4)(1+x^8)(1+x^{16})(1+x^{32})$ &c. $= \frac{P}{1+x}$: facta ergo in Serie eadem substitutione erit $\frac{P}{1+x} = 1 + \alpha x^2 + \beta x^4 + \gamma x^6 + \delta x^8 + \epsilon x^{10} + \zeta x^{12} +$ &c., multiplicetur ergo per $1+x$, eritque $P = 1 + x + \alpha x^2 + \alpha x^3 + \beta x^4 + \beta x^5 + \gamma x^6 + \gamma x^7 + \delta x^8 + \delta x^9 +$ &c., qui valor ipsius $P$ si cum superiori comparetur, habebitur $\alpha = 1; \beta = \alpha; \gamma = \alpha; \delta = \beta; \epsilon = \beta; \zeta = \gamma; \eta = \gamma;$ &c., erunt ergo omnes coëfficientes $= 1$, ideoque productum propositum $P$ evolutum dabit Seriem geometricam $1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7 +$ &c., 329. Cum igitur hic omnes ipsius $x$ Potestates, singulæque semel occurrant, ex forma producti $(1+x)(1+x^2)(1+x^4)$ &c., sequitur, omnem numerum integrum ex terminis progressionis Euleri Introduct. in Anal. infin. parv. Mm gco-